Documentation:CHBE Exam Wiki/Module 6 - Reactive energy Balances

CHBE 241; Exam resources wiki
Chemical and Biological Engineering
	Welcome to the CHBE Exam Resources Wiki! ; This wiki is intended to host past exams; with fully worked-out hints and solutions
Past Exams
	Final Exam 2016W
	Midterm Exam 1 2016W
	Midterm Exam 2 2016W
Problem Sets
	Module 1 - Process Basics
	Module 2 - Reactors
	Module 3 - Separations 1
	Module 4 - Separations 2
	Module 5 - Non-reactive Energy Balances
	Module 6 - Reactive Energy Balances

Question 1

Trichloroethylene is produced in a two-step reaction sequence as shown in the steps below.

Reaction 1: C₂H₄(g) 2Cl₂ (g) → C₂H₂Cl₄(l) + H₂(g) where, ${\hat{H}}_{r}^{\circ} = - 385.76 k J / m o l$
Reaction 2: C₂H₂Cl₄(l) → C₂HCl₃(l) + HCl(g)

The standard heat of formation of tricholoroethylene (liquid) is - 276 2 kJ/mol.

Question 1a

Calculate the standard heat of the second reaction.

Solution

We first need to determine the heat of formation of tetrachloroethane

$\begin{aligned} ({\hat{H}}_{f}^{\circ})_{C_{2} H_{2} C l_{4} (l)} & = ({\hat{H}}_{r 1}^{\circ}) + ({\hat{H}}_{f}^{\circ})_{C_{2} H_{4} (g)} \\ = - 333.48 k J / m o l \\ ({\hat{H}}_{r 2}^{\circ}) & = ({\hat{H}}_{f}^{\circ})_{C_{2} H C l_{3} (l)} + ({\hat{H}}_{f}^{\circ})_{H C l (g)} - ({\hat{H}}_{f}^{\circ})_{C_{2} H_{2} C l_{4} (l)} \\ = - 35.03 k J / m o l \end{aligned}$

Question 1b

Use Hess’s law to calculate the standard heat of the reaction
C₂H₄(g)+ 2Cl₂ (g) → C₂HCl₃(l) + H₂(g) + HCl(g)

Solution

Reaction (1) + Reaction (2)
$\begin{aligned} ({\hat{H}}_{r}^{\circ}) & = ({\hat{H}}_{r 1}^{\circ}) + ({\hat{H}}_{r 2}^{\circ}) \\ = - 385.76 + (- 35.03) \\ = - 420.79 k J / m o l \end{aligned}$

Question 1c

If 450 mol/h of C₂HCl₃ (l) is produced in the reaction of part (b) and the reactants and products are all at 25 C and 1 atm, how much heat is evolved or absorbed in the process?

Solution

Assume that
$\begin{aligned} \dot{Q} & = Δ \dot{H} \\ = 450 m o l / h \cdot (- 420.79 k J / m o l) \\ = - 1.89 \cdot 1 0^{5} k J / h \\ = - 52.60 k W (heat is evolved to the surroundings) \end{aligned}$

Question 2

Ethylene oxide is produced by the oxidation of ethylene under the following reaction:
C₂H₄(g) + 1/2 O₂ (g) → C₂H₄O(g)

Another competing reaction results in oxidation of ethylene to CO.
In a plant, the feed to the reactor contains 2 mol C₂H₄/ mol O₂ while the yield and the conversion in the reactor are 0.8 mol C₂H₄O produced/ mol C₂H₄ consumed and 40% respectively. The outlet stream from the reactor then separates into the following streams:

C₂H₄ and O₂ are recycled back to the reactor
C₂H₄O is sold
CO₂ and H₂O are disposed

Also, the inlet and outlet streams from the reactor are kept at 450°C, while the feed stream and the streams leaving the separator are at 25°C.

Question 2a

Using a basis of 4 moles of ethylene entering the reactor, draw a process flow diagram for the plant.

Solution

Question 2b

Determine the flow rates (molar) and compositions of all the streams.

Solution

With 40% conversion, 1.6 mol of C₂H₄ is consumed. Therefore,
$n_{3} = 2.4 m o l$
With 80% yield,
$\begin{aligned} n_{5} & = (1.6 m o l C_{2} H_{4} c o n s u m e d) \cdot 0.8 \\ = 1.28 m o l \end{aligned}$

C balance on reactor:
$\begin{aligned} 2 (4) & = 2 (2.4) + 2 (1.28) + \cdot n_{6} \\ n_{6} & = 0.64 m o l \end{aligned}$

Since the molar ratio of water to CO₂ produced is 1:1,
$n_{7} = 0.64 m o l$

O balance on the reactor:
$\begin{aligned} 2 (2) & = 2 (n_{4}) + 1.28 + 2 (0.64) + 0.64 \\ n_{4} & = 0.4 m o l \end{aligned}$

Overall C balance:
$\begin{aligned} 2 n_{1} & = 0.64 + 2 (1.28) \\ n_{1} & = 1.6 m o l \end{aligned}$

Overall O balance:
$\begin{aligned} 2 n_{2} & = 2 (0.64) + 0.64 + 1.28 \\ n_{2} & = 1.6 m o l \end{aligned}$

Compositions:
Feed stream: 50% C₂H₄, 50% O₂
Recycle stream: 85.7% C₂H₄,14.2% O₂
Reactor inlet: 66.7% C₂H₄, 33.3% O₂
Reactor outlet: 44.8% C₂H₄, 7.5% O₂, 23.9% C₂H₄O, 11.9% CO₂, 11.9% H₂O

Question 2c

Determine the heat required for the whole process and the reactor alone respectively. Use the following information for C₂H₄O,
$({\hat{H}}_{f}^{\circ}) = - 51 k J / m o l$
$C_{p} [J / (m o l \cdot K)] = - 4.69 + 0.2061 T - 9.995 \cdot 1 0^{- 5} T^{2}$ where T is in kelvin

Solution

By performing energy balance on the process:
$\begin{aligned} Q_{p r o c e s s} & = Δ H \\ = H_{o u t} - H_{i n} \\ = - 583.7 k J \\ Q_{r e a c t o r} & = - 558.7 k J \end{aligned}$

Question 2d

Determine the flow rate (kg/h) and the energy consumption in kW for a production for 5000kg C₂H₄O/day.

Solution

Mass of C₂H₄O produced with initial basis,
$\begin{aligned} m & = 1.28 m o l \cdot 440.5 k g / 1 0^{3} m o l \\ = 0.0564 k g \\ Scale factor & = \frac{5000 k g / d a y}{0.0564 k g} \\ = 88677 / d a y \end{aligned}$

The feed mass at the initial basis,
$\begin{aligned} m_{f^{'} e e d} & = 1.6 m o l \cdot (28.05 g C_{2} H_{4} / m o l) + 1.6 m o l \cdot (32 g O_{2} / m o l) \\ = 96.08 g \end{aligned}$

The fresh feed rate for the production basis of 5000kg/day,
$\begin{aligned} {\dot{m}}_{p r o d u c t i o n} & = 96.08 \cdot 1 0^{- 3} k g \cdot 88677 / d a y \\ = 8520 k g / d a y \end{aligned}$

By performing energy balance on the process:
$\begin{aligned} {\dot{Q}}_{p r o c e s s} & = - 583.7 k J \cdot 88677 / d a y \cdot \frac{1 d a y}{24 h r} \cdot \frac{1 h r}{3600 s} \cdot \frac{1 k W}{1 k J / s} \\ = - 573.4 k W \\ {\dot{Q}}_{r e a c t o r} & = - 599.1 k W \end{aligned}$